dongyin0628 2016-06-25 04:46
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提交表单时如何将值传递到同一页面？ 使用Ajax

Basically, the form has to send data to my database and my database informations should be shown at the same page when user submit the form without refreshing the page. I did something wrong and couldn't find how to fix this. And looked at all the questions but couldn't figure it out. Thanks for the help.

 <div  id="tweetSpace">
      <form id="formTweet" method="post" >
        <textarea id="areaTweet" name="message" rows="2" cols="120" placeholder="Write your tweet here..."></textarea>
        <br>
        <input id="sendTweet" type="submit" value="Send">
      </form>
    </div>

    <div id="txtHint"></div>

<script>
$("#sendTweet").on("submit", function(e){


    var tweet = $('areaTweet').val();
    var update = $('#txtHint');

    $.ajax({
        type: 'POST',
        url: 'tweet2.php',
        data: , tweet,  
        success:function(html){
           update.html(html);
        }
    });
});


</script>

tweet2.php file

<?php 

 include 'connect.php';
  session_start();

$tweet=$_POST['tweet'];
$email =$_SESSION['login_user'];
$sqlr = "INSERT INTO tweets(tweet,member_email) VALUES ('$tweet','$email')";
$rqu = mysqli_query($conn,$sqlr);


$x=0;
$arrayName = array();
$sql= "SELECT tweet FROM tweets WHERE member_email= '$email' "
$rq = mysqli_query($conn,$sql);
while($row = mysqli_fetch_assoc($rq)) {
           $arrayName[$x]  = $row["tweet"];
           $x=$x+1;
}


<?php for($k = 0; $k < $x; $k++) {?>

 <p><?php echo  $arrayName[$k]; ?></p>

<?php } ?


?>
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4条回答 默认 最新

  • doudeng5764 2016-06-25 05:59
    关注

    Here is the working code...Note all changes did in 2 files..

    HTML

    <html>
<head>
    <title>Tweets</title>
</head>

<body>
    <div id="tweetSpace">
      <form id="formTweet" method="post" >
        <textarea id="areaTweet" name="message" rows="2" cols="120" placeholder="Write your tweet here..."></textarea>
        <br>
        <input id="sendTweet" type="button" value="Send">
      </form>
    </div>

    <div id="txtHint"></div>

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.2.2/jquery.min.js"></script>
<script>
  $("#sendTweet").click(function(e){

      var tweet = $('#areaTweet').val();
      var update = $('#txtHint');

      $.ajax({
          type: 'POST',
          data: {'tweet':tweet},
          url: 'tweet2.php',
          success:function(html){
             update.html(html);
          }
      });
  });
</script>

</body>
</html>

    tweet2.php

    <?php 
session_start();
include 'connect.php';
/*
$servername = "localhost";
$username = "root";
$password = "";
$db = "sflow";
$conn = mysqli_connect($servername, $username, $password, $db); 
*/

$tweet = $_POST['tweet'];
$email = /*"sample@s.com";//*/$_SESSION['login_user'];
$sqlr  = "INSERT INTO tweets(tweet,member_email) VALUES ('$tweet','$email')";
$rqu   = mysqli_query($conn,$sqlr);

$x=0;
$arrayName = array();
$sql= "SELECT tweet FROM tweets WHERE member_email= '$email'";
$rq = mysqli_query($conn,$sql);
while($row = mysqli_fetch_assoc($rq)) {
           $arrayName[$x]  = $row["tweet"];
           $x=$x+1;
}

for($k = 0; $k < $x; $k++)
echo '<p>'.$arrayName[$k].'</p>';
?>

    Sample Table

    CREATE TABLE IF NOT EXISTS `tweets` (
  `ID` int(11) unsigned NOT NULL AUTO_INCREMENT,
  `tweet` varchar(255) NOT NULL,
  `member_email` varchar(255) NOT NULL,
  PRIMARY KEY (`ID`),
  UNIQUE KEY `sid_2` (`tweet`),
  KEY `sid` (`tweet`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=3 ;

--
-- Dumping data for table `tweets`
--

INSERT INTO `tweets` (`ID`, `tweet`, `member_email`) VALUES
(1, 'sasa', 's@g.com'),
(2, 'fgfg', 'sample@s.com');
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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